Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef double db;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pa;
typedef unsigned int uint;
typedef unsigned long long ull;
#define w1 first
#define ls (x<<1)
#define w2 second
#define rs (x<<1|1)
#define mp make_pair
#define pb push_back
#define mid ((l+r)>>1)
#define SZ(x) (int((x).size()))
#define All(x) (x).begin(),(x).end()
#define dprintf(...) fprintf(stderr,__VA_ARGS__)
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define rep2(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define per(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)
#define Rep(p,x) for(int (p)=head[(x)];(p);(p)=nxt[(p)])
#define Rep2(p,x) for(int (p)=cur[(x)];(p);(p)=nxt[(p)])
template<class T>void read(T&num){
	num=0;T f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')num=num*10+ch-'0',ch=getchar();
	num*=f;
}
const int maxn=2e5+5;
int n;
int p[maxn],f[maxn],t[maxn],pos[maxn];
bool check(int a,int b,int c){
	if(a<b&&b<c)return 1;
	if(b<c&&c<a)return 1;
	if(c<a&&a<b)return 1;
	return 0;
}
int main(){
	read(n);
	rep(i,1,n)read(p[i]),pos[p[i]]=i;
	t[n]=0;
	per(i,n-1,1){
		if(!t[i+1]){
			if(pos[i]<pos[i+1])t[i]=0;
			else t[i]=1,f[i]=i+1;
		}else{
			if(check(pos[f[i+1]],pos[i],pos[i+1]))t[i]=t[i+1],f[i]=f[i+1];
			else t[i]=t[i+1]+1,f[i]=i+1;
		}
	}
	printf("%d\n",t[1]);
	return 0;
}